MaxWong

$\dfrac{1}{2}+\dfrac{1}{4}\\ =\dfrac{2}{4} + \dfrac{1}{4}\\ = \dfrac{3}{4}$

Same as Mellie's answer.

This could not be the same. 

$\left(\dfrac{a}{b}\right)^{m\cdot n}$

$=\left(\dfrac{a^m}{b^m}\right)^n$

This is correct, this also equals

$\dfrac{a^{mn}}{b^{mn}}$

$\text{This could not be further simplified.}$

y = exp(x) = e^x 

Where e = $1+\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}......$

Which is Euler's number

$\begin{array}{rl}&28-(-43)\\=&28+43\\=&71\end{array}$

$\begin{array}{rl}&-28-(-43)\\=&43-28\\=&15\end{array}$

That is the calculator method.

How to work out 8^(5/3) manually.

$8^{\frac{5}{3}}=(\sqrt[3]{8})^5=2^5=2\cdot 2 \cdot 2 \cdot 2 \cdot 2 = 4 \cdot 2 \cdot 2 \cdot 2 = 8 \cdot 2 \cdot 2 = 16 \cdot 2 = 32\\ \text{Or if you know what's 2^5, the answer is straight-foward-ly 32.}$

See your latest post. Melody replied to your post about the derivative one.

Good job Guest #2

$\dfrac{d \;f(x)}{dx}=1x^{-1}\\ \dfrac{d}{dx} f(x)=\dfrac{1}{x}\\ f(x) = \int \dfrac{dx}{x}\\\quad\;\;\;\;\!=\ln|x|$

$\dfrac{d}{dx} 9\ln|x|=\dfrac{9}{x}$

Melody's way is a good way but I never knew this. Thanks Melody for presenting this method.